Active inductor behaving like resistor

[shaikss]

Hi,

In what scenarios active inductor behaves like a resistor?
I simulated a grounded active inductor circuit by setting the current values I1 = 1u and I0 = 1mA
I sweeped Vdd from 0 to 1.8V.

When I plotted the graph between Impedence seen from M0 and Vdd, I can see only the real part and the imaginary part is zero.
I don't understand why it is behaving so.

M0 is in Saturation region when Vdd is greater than or equal to 4mv and M2 is in Subthreshold region for the entire Vdd range.
Please help me to make the inductor to work with imaginary part of the impedence.
What are the changes required to do in order to see the effect?





Thanks



PS: I have set Vdd as 1.8V and varied i1 and i2 to check the operating regions of the MOS's. When I did so, I observed the below things:

1u <= I1 <= 390u M1 is in Subthreshold
1u <= I1 <= 500u M2 is in Saturation


100u <= I2 <= 1mA M1is in Subthreshold and M2 is in Sat region.

By checking these values, I have set I1 as 350uA and I2 as 600uA.

My intention is to plot real and imaginary part of impedence Vs transconductance.

How to proceed further?


Thanks

 

[BradtheRad]

I haven't worked with active inductor circuits... but I had the idea it arranges a capacitor and op amp to simulate a coil-type inductor.

So I can't say I understand your circuit. It reminds me of a current source rather than something that could would work like an inductor.

But what do I know? So in case your circuit is correct...

Try testing for inductor-like response by applying DC, waiting to see what happens, then switching to zero, and waiting to see what happens.

Don't break the circuit... that's how you get an inductor to generate hi-voltage spikes. I don't know how an active inductor will behave.

The definition of an inductor is that it resists any change in current.

So after you apply DC, you should observe a delay while current builds.

When you switch to zero, you should get continued current flow in the loop for a short time afterward.

Your component values will reflect a time constant, corresponding to the formula L/R which is used for a normal inductor.

Then progress to AC square waves. Current flow should lag after each reversal of polarity.

Then you can progress to sine waves, and observe whether reactance turns out to be what's expected, etc.

 

[krishnaengr]

hi sir
i am also trying with the same circuit .
for this circuit whether i have to give any input voltage.
then whether you found any value for the i1 and i2.please help in this values